Problem: Let $g(x)=\begin{cases} \sin(x)&\text{for }x<0 \\\\ x^2&\text{for }x\geq 0 \end{cases}$ Is $g$ continuous at $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: For $g$ to be continuous at $x=0$, we need $\lim_{x\to 0}g(x)$ and $g(0)$ to exist and be equal. Since $0\geq 0$, the rule that applies to $x=0$ is $x^2$. So $g(0)=0^2=0$. Now let's analyze $\lim_{x\to 0}g(x)$. Finding $\lim_{x\to 0^{ +}}g(x)$ For $x$ -values larger than $0$, the appropriate rule for $g(x)$ is $x^2$. Since $x^2$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ +}}g(x) \\\\ &=\lim_{x\to 0^{ +}}x^2 \gray{x^2\text{ is the rule for }x>0} \\\\ &=(0)^2 \gray{x^2\text{ is continuous at }x=0} \\\\ &=0 \end{aligned}$ Finding $\lim_{x\to 0^{ -}}g(x)$ For $x$ -values smaller than $0$, the appropriate rule for $g(x)$ is $\sin(x)$. Since $\sin(x)$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ -}}g(x) \\\\ &=\lim_{x\to 0^{ -}}[\sin(x)] \gray{\sin(x)\text{ is the rule for }x<0} \\\\ &=\sin(0) \gray{\sin(x)\text{ is continuous at }x=0} \\\\ &=0 \end{aligned}$ Conclusion We found that: $\lim_{x\to 0^{ +}}g(x)=\lim_{x\to 0^{ -}}g(x)=g(0)=0$ Since the one-sided limits are both equal to $g(0)$, we can determine that the two-sided limit $\lim_{x\to 0}g(x)$ is also equal to $g(0)$, and $g$ is continuous at $x=0$.